Lebesgue measure and integral

  • #1
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Hello. I have problem with this integral :
[tex]\lim_{n \to \infty } \int_{\mathbb{R}^+} \left( 1+ \frac{x}{n} \right) \sin ^n \left( x \right) d\mu_1[/tex] where ## \mu_1## is Lebesgue measure.
 
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  • #2
Svein
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Since the integrand is continuous in x for each n>0, you can safely use the standard Riemann integral. So -set [itex]I_{n}=\int_{0}^{\infty}(1+\frac{x}{n})\sin^{n}(x)dx=\int_{0}^{\infty}\sin^{n}(x)dx+\int_{0}^{\infty}\frac{x}{n}\sin^{n}(x)dx [/itex]. The first integral is trivial, use partial integration on the second.
 
  • #3
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@Svein: I'm quite sure these integrals are not well-defined, especially for (but not limited to) even n.
 
  • #4
Svein
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@Svein: I'm quite sure these integrals are not well-defined, especially for (but not limited to) even n.
I agree that a proof might be a little hairy, but it depends on fixing an upper limit (K) in the integrals, then letting n→∞ (use an ε, n argument). These limits are independent of K.

Another thought: You could also argue that as n→∞ sinnx →o a. e.
 
  • #5
Erland
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But the integrals
[tex]\int_{\mathbb{R}^+} \left( 1+ \frac{x}{n} \right) \sin ^n \left( x \right) d\mu_1[/tex] where ## \mu_1## is Lebesgue measure
diverge for each ##n## (to ##+\infty## if ##n## is even, not even to that if ##n## is odd), and it makes no sense to take the limit of a sequence whose elements are not defined. The limit expression is therefore meaningless.
 
  • #6
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Exactly. If we could take the limit for n to infinity first it would work, but that is not the problem statement.
 
  • #7
Svein
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Exactly. If we could take the limit for n to infinity first it would work, but that is not the problem statement.
Well, somehow it is.
  1. Any integral of the type [itex] \int_{0}^{\infty}f(x)d\mu[/itex] is in reality a limit process: [itex]\lim_{A\rightarrow \infty}\int_{0}^{A}f(x)d\mu [/itex]
  2. I propose that [itex]\lim_{n\rightarrow \infty}\int_{0}^{A}(1+\frac{x}{n})\sin^{n}(x) d\mu = 0 [/itex] independent of A.
  3. The reason for this is that [itex]\lim_{n\rightarrow \infty}\sin^{n}(x) [/itex] is 0 everywhere except for [itex]x=m\cdot \pi [/itex], m any integer.
  4. Thus the limit in point 2 is proved.
  5. As A→∞, m does likewise. But still [itex]\lim_{n\rightarrow \infty}\sin^{n}(x) [/itex] is 0 everywhere except for a countable number of points.
 
  • #8
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@Svein: You cannot do this. The problem statement can be expressed as
$$\lim_{n\rightarrow \infty} \lim_{A\rightarrow \infty} \int_{0}^{A}(1+\frac{x}{n})\sin^{n}(x) d\mu$$
This is not the same as
$$\lim_{A\rightarrow \infty} \lim_{n\rightarrow \infty} \int_{0}^{A}(1+\frac{x}{n})\sin^{n}(x) d\mu$$
You try to evaluate the second expression. To do that you have to show that the two expressions are equal. Often you can exchange the order of limits, but this here is an example where you cannot.

An easier example would be:
$$\lim_{n\rightarrow \infty} \lim_{c\rightarrow 0} \int_{c}^{1} max(n-xn^2,0) dx = 1/2$$
$$\lim_{c\rightarrow 0} \lim_{n\rightarrow \infty} \int_{c}^{1} max(n-xn^2,0) dx = 0$$
 
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  • #9
Svein
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I admit that my specialty lies in another field (complex function algebras - 50 years ago), but I have checked my copy of Royden: Real Analysis. There are several theorems that I think can be applied, such as the Monotone Convergence Theorem (observing that [itex]\left\lvert \int f(x)\sin^{n}(x)d\mu \right\rvert\leq \int \left\lvert f(x)\right\rvert\ \left\lvert \sin^{n}(x)\right\rvert\ d\mu [/itex]).
 
  • #10
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That doesn't help if both sides are undefined for every n.
 
  • #11
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Thanks everyone for help.
 

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